The general solution of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given differential equation is $e^{x} dy + (ye^{x} + 2x) dx = 0$. Dividing by $dx$,we get $e^{x} \frac{dy}{dx} + ye^{x} + 2x = 0$. Dividing by

Vedclass · Accepted Answer

$ye^{x} + x^{2} = C$

Vedclass · Answer

(D) The given differential equation is $e^{x} dy + (ye^{x} + 2x) dx = 0$. Dividing by $dx$,we get $e^{x} \frac{dy}{dx} + ye^{x} + 2x = 0$. Dividing by $e^{x}$,we get $\frac{dy}{dx} + y = -2x e^{-x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = -2x e^{-x}$. The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int 1 dx} = e^{x}$. The general solution is given by $y(I.F.) = \int (Q 	imes I.F.) dx + C$. Substituting the values,$y e^{x} = \int (-2x e^{-x} \cdot e^{x}) dx + C$. $y e^{x} = \int -2x dx + C$. $y e^{x} = -x^{2} + C$. Therefore,$y e^{x} + x^{2} = C$.

The general solution of the differential equation $e^{x} dy + (ye^{x} + 2x) dx = 0$ is

Similar Questions

An integrating factor for the differential equation $(1 + y^2)dx - (\tan^{-1} y - x)dy = 0$ is:

The general solution of the differential equation $y+\cos x(\frac{dy}{dx})-\cos^2 x=0$ is

The curve satisfying the differential equation $y \, dx - (x + 3y^2) \, dy = 0$ and passing through the point $(1, 1)$ also passes through the point

Observe the following statements:
$I$. If $dy+2xy dx=2e^{-x^2} dx$,then $ye^{x^2}=2x+c$
$II$. If $ye^{x^2}-2x=c$,then $dx=\frac{dy}{2e^{-x^2}-2xy}$
Which of the following is a correct statement?

For the primitive integral equation $ydx + y^2dy = xdy$ ; $x \in R$,$y > 0$,$y = y(x)$,$y(1) = 1$,then $y(-3)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module